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[SOWHAT0819]

3z/(z-2)^2, anti-causal inverse transform

 

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To find the anti-causal inverse transform of the given function \( \frac{3z}{(z-2)^2} \), we need to first E×ρréšš it in terms of a power series.

Given function: \( F(z) = \frac{3z}{(z-2)^2} \)

Let's rewrite the function in partial fraction form:

\( F(z) = 3 \cdot \frac{1}{(z-2)^2} \)

Expanding the terms using the formula for the inverse Z-transform of \( \frac{1}{(z-a)^n} \), we get:

\( f[n] = a^{n-1} u[n-1] \)

Applying this to our function, we have:

\( f[n] = 2^{n-1} u[n-1] \)

Therefore, the anti-causal inverse Z-transform of \( \frac{3z}{(z-2)^2} \) is \( 2^{n-1} u[n-1] \).